Question

Define Electrical conductivity. Obtain the expression of electrical conductivity of a conductor in terms of number density and relaxation time of free electrons.
(b) Explain qualitative change in resistivity of a conductor with temperature using expression obtained in (a).

Hint:

sigma = n e^2 tau / m. Resistivity rho = m/(n e^2 tau): tau falls when T rises, so rho rises.

Answer 1

Step 1: Definition of electrical conductivity:
Electrical conductivity \( \sigma \) of a material is the reciprocal of its resistivity, \( \sigma = \dfrac{1}{\rho} \). In other words, \( \sigma = \dfrac{J}{E} \), the current density produced per unit electric field. Its SI unit is \( \text{S m}^{-1} \) (or \( \Omega^{-1}\text{m}^{-1} \)).

Step 2: Drift velocity in terms of relaxation time:
Free electrons in a conductor undergo collisions with the lattice ions. The average time between two collisions is the relaxation time \( \tau \).
Under a field \( E \), each electron gets an acceleration \( a = \dfrac{eE}{m} \). It is thrown to a random velocity at each collision, so the average drift velocity is \( v_d = a\tau = \dfrac{eE\tau}{m} \).

Step 3: Current density and conductivity:
If \( n \) is the number of free electrons per unit volume, the current density is \( J = n e v_d = \dfrac{n e^2 \tau}{m}\,E \).
Compare with \( J = \sigma E \). Then \( \sigma = \dfrac{n e^2 \tau}{m} \).

Step 4: Part (b): resistivity and temperature:
Resistivity is \( \rho = \dfrac{1}{\sigma} = \dfrac{m}{n e^2 \tau} \).
When the temperature of a metal rises, the lattice ions vibrate more strongly. The electrons collide more often, so the relaxation time \( \tau \) decreases.
The number density \( n \) of free electrons in a metal stays almost constant with temperature. So \( \rho \) increases when \( \tau \) decreases.
Thus the resistivity of a conductor increases with temperature.

Final Answer:
(a) \( \sigma = \dfrac{ne^2\tau}{m} \). (b) \( \rho = \dfrac{m}{ne^2\tau} \) rises with temperature because \( \tau \) falls while \( n \) is nearly constant.